Network analysis is the general name given to
certain specific techniques which can be used
for the planning, management and control of
projects.
 Use of nodes and arrows:Arrows  An arrow leads from tail to head
directionally


Indicate ACTIVITY, a time consuming effort that is
required to perform a part of the work.
•
NODE:A
node is represented by a circle
- Indicate EVENT, a point in time where one or
more activities start and/or finish.
•
Activity:–
–
–
•
A task or a certain amount of work required in the
project
Requires time to complete
Represented by an arrow
Dummy Activity:–
–
Indicates only precedence relationships
Does not require any time of effort.



Event:


Signals the beginning or ending of an activity
Designates a point in time
Represented by a circle (node)

Shows the sequential relationships among
activities using nodes and arrows
Network:-
Activity-on-node (AON):nodes represent activities, and arrows show precedence
relationships

Activity-on-arrow (AOA):arrows represent activities and nodes are events for points
in time
SITUATIONS IN NETWORK DIAGRAM:B
A must finish before either B or C
can start.
C
A
Both A and B must finish before C
can start.
C
B
A
Both A and B must finish before
either of C or D can start.
C
D
B
A must finish before B can start
both A and C must finish before
D can start.
B
A
Dummy
C
D
illustration of network analysis of a minor
redesign of a product and its associated
packaging.
The key question is: How long will it take to
complete this project ?


Path


Critical Path


A connected sequence of activities leading from the
starting event to the ending event
The longest path (time); determines the project
duration
Critical Activities

All of the activities that make up the critical path.
Forward Pass:
Earliest Start Time (ES)



earliest time an activity can start
ES = maximum EF of immediate predecessors
Earliest finish time (EF)


earliest time an activity can finish
earliest start time plus activity time
EF= ES+t
Backward Pass:
Latest Start Time (LS)
Latest time an activity can start without delaying critical
path time
LS= LF - t

Latest finish time (LF)
latest time an activity can be completed without
delaying critical path time
LS = minimum LS of immediate predecessors





Draw the CPM network
Analyze the paths through the network
Determine the float for each activity
 Compute the activity’s float
float = LS - ES = LF - EF
 Float is the maximum amount of time that this
activity can be delay in its completion before it
becomes a critical activity, i.e., delays completion of
the project
Find the critical path is that the sequence of activities
and events where there is no “slack” i.e.. Zero slack
 Longest path through a network
Find the project duration is minimum project

CPM Network:f, 15
h, 9
g, 17
a, 6
i, 6
b, 8
d, 13
c, 5
e, 9
j, 12

ES and EF Times:f, 15
a, 6
i, 6
0 6
b, 8
0 8
0
h, 9
g, 17
5
d, 13
c, 5
e, 9
j, 12

ES and EF Times:6
h, 9
g, 17
a, 6
0
21
f, 15
6
23
i, 6
6
b, 8
0
0
8
5
j, 12
d, 13
c, 5
8
21
e, 9
5
14

ES and EF Times:6
21
f, 15
21 30
0
6
h, 9
g, 17
a, 6
6
0
23
i, 6
8
23 29
b, 8
d, 13 8
0
5
21
j, 12
21 33
c, 5
e, 9
Project’s EF = 33
5
14

LS and LF Times:6
0
0
0
8
0
b, 8
5
0
h, 9
g, 17
a, 6
0
0
21 30
24 33
f, 15
6
0
0
0
21
0
6
0
23
0
i, 6 23 29
27 33
d, 13
8
0
21
0
j, 12
21 33
21 33
c, 5
e, 9
5
0
14
0

LS and LF Times:6 21
18 24
0
4
f, 15
6
10
0
7
5
12
h, 9
6 23 g, 17
10 27
a, 6
0
0
b, 8
21 30
24 33
8
8
8
8
21
21
23 29
27 33 i, 6
d, 13
j, 12
21 33
21 33
c, 5
e, 9
5 14
12 21

FLOAT:-
6
9
03
21
24
f, 15
0 6
4 10
a, 6
3
3
h, 9
g, 17
6 23
04
10 27
4
0
0
b, 8
0
7
0
7
5
12
8
8
0
8
8
21
21
d, 13
21 30
24 33
23 29 i, 6
27 33
j, 12
0
c, 5
e, 9
7
5 14
12 21
21 33
21 33

Critical Path:f, 15
h, 9
g, 17
a, 6
i, 6
b, 8
d, 13
c, 5
e, 9
j, 12


PERT is based on the assumption that an activity’s
duration follows a probability distribution instead
of being a single value
Three time estimates are required to compute the
parameters of an activity’s duration distribution:
 pessimistic time (tp ) - the time the activity
would take if things did not go well
 most likely time (tm ) - the consensus best
estimate of the activity’s duration
 optimistic time (to ) - the time the activity would
take if things did go well
te = a+4m+b
6





Draw the network.
Analyze the paths through the network and find
the critical path.
The length of the critical path is the mean of the
project duration probability distribution which is
assumed to be normal
The standard deviation of the project duration
probability distribution is computed by adding
the variances of the critical activities (all of the
activities that make up the critical path) and
taking the square root of that sum
Probability computations can now be made using
the normal distribution table.
Determine probability that project is completed
within specified time
Z=
x-





where  = tp = project mean time
 = project standard mean time
x = (proposed ) specified time
Probability
Z
 = tp
x
Time





Useful at many stages of project management
Mathematically simple
Give critical path and slack time
Provide project documentation
Useful in monitoring costs
•
•
•
•
How long will the entire project take to be
completed? What are the risks involved?
Which are the critical activities or tasks in the
project which could delay the entire project if they
were not completed on time?
Is the project on schedule, behind schedule or
ahead of schedule?
If the project has to be finished earlier than
planned, what is the best way to do this at the least
cost?



Parallel paths-identifying a single path is difficult
when there are parallel paths with similar duration.
Time consuming-critics note that it takes too much
time to identify all activities and inter-relate them to
get multiple projects paths.
First time projects-CPM is not suitable if projects
cannot be broken down into discrete activities with
known completion times.
PRESENTED BY:BHUPENDRA SINGH SHEKHAWAT
ANKIT VINOD AGRAWAL
BHANU MATHUR
AMIT SINGAL
AKANSHA CHOUDHARY
KAMAL KANT
AKASH GARG
MOHIT SHARMA
ANKIT BAJORIA
MAYANK AGRAWAL